Animated Solution for Physics - Oscillations: When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y(t)=y_0{\sin }^2\omega t$, where $y$ is measured from the lower end of unstretched spring. Then $\omega$ is

When you first look at the equation of motion $y(t)=y_0{\sin }^2\omega t$, it might seem a bit alien. Standard Simple Harmonic Motion (SHM) is always described by a linear sine or cosine function, like $y(t)=A\sin (\omega t+\varphi )$. The presence of the square on the sine function is a classic trap set by examiners to test your mathematical agility and your deep understanding of physical parameters.

To make sense of this, we must first linearize the equation. We can call upon a trusty double-angle trigonometric identity:

Substituting this into our given displacement equation, we get:

Expanding this expression gives us a much more familiar form:

Now, let's rearrange this equation to represent the displacement from a central point:

By comparing this with the standard equation of an oscillator $Y=-A\cos ({\omega }^{\prime }t)$, we can extract a wealth of physical information.

First, the mean (equilibrium) position of the oscillation is exactly at $y_{eq}=\frac{y_0}{2}$. Second, the amplitude of the oscillation is $A=\frac{y_0}{2}$. Most importantly, the true angular frequency of the system is not $\omega$, but rather ${\omega }^{\prime }=2\omega$. This is the critical insight that unlocks the rest of the problem!

We know that for any spring-mass system, the true angular frequency is determined by the spring constant $k$ and the mass $m$:

Since we established that ${\omega }^{\prime }=2\omega$, we can write:

To proceed, we need to find the spring constant $k$. Let's analyze the forces at the equilibrium position. At $y_{eq}=\frac{y_0}{2}$, the net force on the block is zero. The upward restoring force of the spring perfectly balances the downward pull of gravity.

Rearranging this to solve for $k$, we get:

Now, we substitute this expression for $k$ back into our angular frequency equation:

Notice how beautifully the mass $m$ cancels out, leaving us with:

Finally, dividing both sides by 2, we isolate $\omega$:

This perfectly matches option (c).

The 10-Second Alternative Method

If you want to save precious time in a competitive exam, you can bypass the equilibrium analysis entirely by using initial conditions!

At time $t=0$, the displacement is $y(0)=0$. The spring is completely unstretched, meaning the spring force is zero. The only force acting on the block is gravity, so its initial acceleration must be $g$.

Now, let's find the acceleration mathematically by differentiating the displacement equation twice:

At $t=0$, the mathematical acceleration is:

Equating the physical acceleration to the mathematical acceleration:

Boom! You arrive at the exact same answer in a fraction of the time. This highlights the power of understanding the physical meaning behind the math.